In the figure below, stick figure O (the object) stands on the common central ax

Question

In the figure below, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses. Lens 1 is mounted within the boxed region closer to O, which is at object distance p1. Lens 2 is mounted within the farther boxed region, at distance d. The table refers to the focal distances f_1 and f_2 for lens 1 and lens 2 respectively, the image distance i_2 for the final mage produced by lens 2 (the final image produced by the system) and the overall lateral magnification m for the system. All distances are in centimeters. Fill in the missing information, including signs. Describe the image. (Select all that apply.) real virtual upright inverted on the same side of lens 2 as object O on the opposite side of lens 2

Explanation / Answer

Since lens 1 is converging, f1 = +17 cm, and we find the image distance to be

1/f1= 1/p1 +1/i1

i1 = p1f1/(p1-f1)

i1 = 11*17/(11-17)

i1 = -31.1667 cm

This serves as an “object” for lens 2 (which has f2 = +7.6 cm) with an object distance given by p2 = d – i1 = 41.16 cm.

90

i2 = p2f2/(p2-f2)

i2 = 9.32 cm

(b) The overall magnification is M = m1m2 = (-i1 / p1)(-i2 / p2 )

M = i1i2/p1p2

M = – 0.64

(c) The fact that the (final) image distance is negative means the image is virtual (V).

(d) The fact that the magnification is a negative value means the image is inverted (I).

(e) The image is on the same side as the object (relative to lens 2).

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